A wheel of radius 0.2 mathrm~m$0.2 \mathrm{~m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 mathrm~N$10 \mathrm{~N}$ as shown in figure. The established torque produces an angular acceleration of 2 mathrmrad / mathrms^2$2 \mathrm{rad} / \mathrm{s}^{2}$ . Moment of inertia of the wheel is ________ kg m^2$m^2$ . (Acceleration due to gravity = 10mathrmm / mathrms^2$= 10\mathrm{m} / \mathrm{s}^2$)

### Related Formula The definition of angular momentum vector vecL$\vec{L}$ relative to the origin is: vecL = vecr times vecp = m(vecr times vecv) $$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v}) $$ In scalar form for a horizontal speed component at a maximum altitude H$H$: L = m v_x H$L = m v_{x} H$ ### Core Logic At maximum height, the vertical speed component drops to zero, and the projectile travels entirely horizontally along the curve profile as shown in

: v_x = v_0 cos 45^circ = fracv_0sqrt2$$v_{x} = v_{0} \cos 45^{\circ} = \frac{v_{0}}{\sqrt{2}}$$ H = fracv_0^2 sin^2 45^circ2g = fracv_0^24g$$H = \frac{v_{0}^{2} \sin^{2} 45^{\circ}}{2g} = \frac{v_{0}^{2}}{4g}$$ ### Step 1: Calculating Magnitude and Vector Direction Evaluate the horizontal vector cross components: L = m left(fracv_0sqrt2 ight) left(fracv_0^24g ight) = fracmv_0^34sqrt2g$$L = m \left(\frac{v_{0}}{\sqrt{2}} ight) \left(\frac{v_{0}^{2}}{4g} ight) = \frac{mv_{0}^{3}}{4\sqrt{2}g}$$ Using the right-hand rule, vecr$\vec{r}$ points into quadrant-1 while velocity points towards +hati$+\hat{i}$. Therefore, vecr times vecv$\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative hatk$\hat{k}$ orientation (along the negative z-axis). ### Pattern Recognition At peak height, always map vecL$\vec{L}$ using m cdot v_textpeak cdot y_textmax$m \cdot v_{\text{peak}} \cdot y_{\text{\max}}$. This scalar form simplifies the calculation significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula The linear acceleration a$a$ for pure rolling motion down an incline profile is: a = fracgsintheta1 + fracImr^2$$a = \frac{g\sin\theta}{1 + \frac{I}{mr^{2}}}$$ ### Core Logic For a uniform solid cylinder, the moment of inertia around its central axis is: I = frac12mr^2 implies fracImr^2 = frac12$$I = \frac{1}{2}mr^{2} \implies \frac{I}{mr^{2}} = \frac{1}{2}$$ ### Step 1: Calculating Acceleration Substitute theta = 45^circ$\theta = 45^{\circ}$ and the cylinder inertial factor into the formula : a = fracgsin 45^circ1 + frac12 = fracfracgsqrt2frac32$$a = \frac{g\sin 45^{\circ}}{1 + \frac{1}{2}} = \frac{\frac{g}{\sqrt{2}}}{\frac{3}{2}}$$ a = frac2g3sqrt2 = fracsqrt2g3$$a = \frac{2g}{3\sqrt{2}} = \frac{\sqrt{2}g}{3}$$ ### Pattern Recognition Solid cylinder rolls with acceleration matching frac23 gsintheta$\frac{2}{3} g\sin\theta$. Since sin 45^circ = frac1sqrt2$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to fracsqrt2g3$\frac{\sqrt{2}g}{3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques: sum tau_textpivot = 0 implies sum (m_i cdot g cdot x_i) = 0$$\sum \tau_{\text{pivot}} = 0 \implies \sum (m_i \cdot g \cdot x_i) = 0$$ ### Core Logic The rod is uniform, meaning its mass (250text g$250\text{ g}$) acts exactly at its geometric center of mass, the 50text cm$50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40text cm$40\text{ cm}$ mark . Calculate the relative lever arms from the pivot [cite: 775, 776, 777]: * 400text g$400\text{ g}$ mass at 10text cm$10\text{ cm}$ mark: lever arm = 40 - 10 = 30text cm$= 40 - 10 = 30\text{ cm}$ (counter-clockwise) * 250text g$250\text{ g}$ rod mass at 50text cm$50\text{ cm}$ mark: lever arm = 50 - 40 = 10text cm$= 50 - 40 = 10\text{ cm}$ (clockwise) * Unknown mass M$M$ at 90text cm$90\text{ cm}$ mark: lever arm = 90 - 40 = 50text cm$= 90 - 40 = 50\text{ cm}$ (clockwise) Setting up the torque balance equation: 400 times 30 = (250 times 10) + (M times 50)$$400 \times 30 = (250 \times 10) + (M \times 50)$$ 12000 = 2500 + 50M$$12000 = 2500 + 50M$$ 50M = 9500 implies M = frac950050 = 190text g$$50M = 9500 \implies M = \frac{9500}{50} = 190\text{ g}$$ ### Step 1: Visual Context The structural layout of forces acting on the balanced rod system is shown below:

### Pattern Recognition Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula vectau = vecr times vecF$$\vec{\tau} = \vec{r} \times \vec{F}$$ ### Core Logic Given position vector vecr = hati + hatj + hatk$\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force vecF = hati - hatj + hatk$\vec{F} = \hat{i} - \hat{j} + \hat{k}$: vectau = left| beginarrayccc hati & hatj & hatk \\ 1 & 1 & 1 \\ 1 & -1 & 1 endarray right|$$\vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right|$$ ### Step 1: Isolate z-component tau_z = hatk(1(-1) - 1(1)) = -2hatk$$\tau_z = \hat{k}(1(-1) - 1(1)) = -2\hat{k}$$ The absolute magnitude of the torque component in the z$z$-direction equals 2 mathrm~N cdot m$2 \mathrm{~N \cdot m}$. ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Position vector of the Centre of Mass (COM): vecr_textCOM = fracm_1vecr_1 + m_2vecr_2 + m_3vecr_3m_1 + m_2 + m_3$$\vec{r}_{\text{COM}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$$ ### Core Logic Assign coordinates to the three masses (m_1=m_2=m_3=2M$m_1=m_2=m_3=2M$): - Origin mass: vecr_1 = 0hati + 0hatj$\vec{r}_1 = 0\hat{i} + 0\hat{j}$ - X-axis mass: vecr_2 = 4hati + 0hatj$\vec{r}_2 = 4\hat{i} + 0\hat{j}$ - Y-axis mass: vecr_3 = 0hati + 4hatj$\vec{r}_3 = 0\hat{i} + 4\hat{j}$ Substitute these into the COM formula: vecr_textCOM = frac2M(0) + 2M(4hati) + 2M(4hatj)2M + 2M + 2M = frac8Mhati + 8Mhatj6M = frac43hati + frac43hatj$$\vec{r}_{\text{COM}} = \frac{2M(0) + 2M(4\hat{i}) + 2M(4\hat{j})}{2M + 2M + 2M} = \frac{8M\hat{i} + 8M\hat{j}}{6M} = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j}$$ ### Step 1: Calculate Position Vector Magnitude |vecr_textCOM| = sqrtleft(frac43right)^2 + left(frac43right)^2 = frac4sqrt23$$|\vec{r}_{\text{COM}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$ Matching this directly with the given template frac4sqrt2x$\frac{4\sqrt{2}}{x}$ shows that x = 3$x = 3$. ### Pattern Recognition Since the mass layout is completely symmetric along both right-angle legs, the COM coordinates are identical (x_textCOM = y_textCOM$x_{\text{COM}} = y_{\text{COM}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion